Question

Evaluate

$\begin{array}{c}lim\\ x\to a\end{array}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\left(from\frac{0}{0}\right)$

Answer 1

We have,

${lim}_{x\to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\left(\frac{0}{0}form\right)$

Now, Applying L’ Hospital rule and we get,

${lim}_{x\to a}\frac{\frac{1}{2\sqrt{a+2x}}\times 2-\frac{1}{2\sqrt{3x}}\times 3}{\frac{1}{2\sqrt{3a+x}}-2\times \frac{1}{2\sqrt{x}}}$

Taking limit and we get,

$\frac{\frac{1}{2\sqrt{a+2a}}\times 2-\frac{1}{2\sqrt{3a}}\times 3}{\frac{1}{2\sqrt{3a+a}}-2\times \frac{1}{2\sqrt{a}}}$

$\Rightarrow \frac{\frac{1}{\sqrt{3a}}-\frac{3}{2\sqrt{3a}}}{\frac{1}{2\sqrt{4a}}-\frac{1}{\sqrt{a}}}$

$\Rightarrow \frac{\frac{2-3}{2\sqrt{3a}}}{\frac{1}{4\sqrt{a}}-\frac{1}{\sqrt{a}}}$

$\Rightarrow \frac{\frac{2-3}{2\sqrt{3a}}}{\frac{1-4}{4\sqrt{a}}}$

$\Rightarrow \frac{\frac{-1}{2\sqrt{3}}}{\frac{-3}{4}}$

$\Rightarrow \frac{-4}{-6\sqrt{3}}$

$\Rightarrow \frac{2}{3\sqrt{3}}\text{on}\text{rationalize}$

$\Rightarrow \frac{2\sqrt{3}}{9}$

Hence, this is the answer.